Got 97.5 percentile in jee mains general category what nit core branches can I get...

Hello Yoyo

Congrats for 97.5 percentile in JEE mains 2019.

This is very good percentile but we cant say that in which NIT you will get admission but we can provide you a list so that you can apply for that after the released cutt from NTA in April.

List of NIT's in which you can try are

• National Institute of Technology, Jamshedpur, Jharkhand

• National Institute of Technology, Kurukshetra, Haryana

• Visvesvaraya National Institute of Technology, Nagpur, Maharashtra

• National Institute of Technology, Patna, Bihar

• National Institute of Technology, Rourkela, Orissa

• National Institute of Technology, Srinagar, Jammu & Kashmir

Sardar Vallabhbhai National Institute of Technology, Surat, Gujarat

National Institute of Technology, Raipur, Chhattisgarh

National Institute of Technology, Agartala, Tripura

Indian Institute of Information Technology, Allahabad, Uttar Pradesh

Some Core Branches you can apply for -->

• Civil Engineering
  
• Chemical Engineering
  
• Biotechnology
  
• Computer Science
  
• Mechanical
  
• Electronics and telecommunication
  
• Chemical
  

I hope you understand my point.

thank you and GOOD LUCK!!!

Hello. Though it is not possible to tell your exact rank but probably it will be somewhere between 26,846 - 30,846. It is possible for you to get a good nit. But the percentile and ranks may change after the April exam. So to know your exact rank you need to wait till the release of the overall merit list by NTA.

Meanwhile you can check the below link to know more about cutoff for various colleges in the previous years.

https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff-for-top-nits/amp

Hope this helps. For any other questions feel free to comment down below.

Hello Aspirant

Firstly congrats for your percentile!!

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Secondly it is difficult to predict the college because the official cutoff and results are not released by the national testing agency NTA. It was released on the month of April after the second jEE mains exam was conducted.

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Use this formula to predict the rank:

Rank Formula = (100 - your total score) X 874469 /100.

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Your rank by this formula is 21861 out of 874469 aspirants. But this rank may differ from the official results which were released on the month of April.



If you want to know the cutoff for top Nits I will provide the link below from careers 360 Once go through the link you will get an idea:

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https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff-for-top-nits/amp

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If you have any questions further let me know.

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Thank you!!!!